Is ${908335}$ divisible by $9$ ?
Solution: A number is divisible by $9$ if the sum of its digits is divisible by $9$ . [ Why? First, we can break the number up by place value: $ \begin{eqnarray} {908335}= &&{9}\cdot100000+ \\&&{0}\cdot10000+ \\&&{8}\cdot1000+ \\&&{3}\cdot100+ \\&&{3}\cdot10+ \\&&{5}\cdot1 \end{eqnarray} $ Next, we can rewrite each of the place values as $1$ plus a bunch of $9$ s: $ \begin{eqnarray} {908335}= &&{9}(99999+1)+ \\&&{0}(9999+1)+ \\&&{8}(999+1)+ \\&&{3}(99+1)+ \\&&{3}(9+1)+ \\&&{5} \end{eqnarray} $ Now if we distribute and rearrange, we get this: $ \begin{eqnarray} {908335}= &&\gray{9\cdot99999}+ \\&&\gray{0\cdot9999}+ \\&&\gray{8\cdot999}+ \\&&\gray{3\cdot99}+ \\&&\gray{3\cdot9}+ \\&& {9}+{0}+{8}+{3}+{3}+{5} \end{eqnarray} $ Any number consisting only of $9$ s is a multiple of $9$ , so the first five terms must all be multiples of $9$ That means that to figure out whether the original number is divisible by $9 $ , all we need to do is add up the digits and see if the sum is divisible by $9$ . In other words, ${908335}$ is divisible by $9$ if ${ 9}+{0}+{8}+{3}+{3}+{5}$ is divisible by $9$ Add the digits of ${908335}$ $ {9}+{0}+{8}+{3}+{3}+{5} = {28} $ If ${28}$ is divisible by $9$ , then ${908335}$ must also be divisible by $9$ ${28}$ is not divisible by $9$, therefore ${908335}$ must not be divisible by $9$.